{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Statement\n\nLet $a \u003c b$, let $K \\colon [a, b] \\times [a, b] \\to \\C$ be a [[continuous kernel|ZM5G144PZG]], and let $M = \\max \\abs{K}$.\nThen for every $n \\ge 1$ and all $x_1, \\dots, x_n, y_1, \\dots, y_n \\in [a, b]$,\n$$\\abs{\\det[K(x_i, y_j)]_{i,j=1}^{n}} \\le n^{n/2} M^n.$$\nConsequently, the $n$-th coefficient $c_n = \\frac{1}{n!} \\int_{[a, b]^n} \\det[K(x_i, x_j)] \\, dx$ of the [[Fredholm determinant|PPX0AK81QV]] satisfies\n$$\\abs{c_n} \\le \\frac{n^{n/2} \\, M^n (b - a)^n}{n!},$$\nand the series $\\sum_n (-1)^n c_n \\lambda^n$ converges for every $\\lambda \\in \\C$.\n\n## Remarks\n\nThe first bound is [[Hadamard's inequality|CGDN1X32NQ]] for the matrix $[K(x_i, y_j)]$: each column has $n$ entries of modulus at most $M$, so its Euclidean norm is at most $\\sqrt{n} M$.\nIntegrating over $[a, b]^n$ gives the bound on $c_n$.\nBy Stirling's formula $n! \\ge (n/e)^n$, so $\\abs{c_n \\lambda^n} \\le \\left(e M (b - a) \\abs{\\lambda}\\right)^n / n^{n/2}$, and the $n$-th root of this tends to $0$.\nHence the radius of convergence is infinite, and $D$ is an [[entire function|0G4F0XB9X2]].\n\nThe same argument, applied to matrices of size $n + k$, shows that the series defining the [[Fredholm minors|HP5VJ9EBST]] converge for every $\\lambda$, uniformly in the fixed points.\n\nThe crude bound $\\abs{\\det} \\le n! \\, M^n$, from expanding the determinant into $n!$ products, is not enough: it only gives $\\abs{c_n} \\le (M (b - a))^n$ and hence convergence for $\\abs{\\lambda} \u003c 1 / (M (b - a))$.\nThe factor $n^{n/2}$ instead of $n!$ is what makes the whole theory work for every $\\lambda$.\n\n## Explanations\n\n### L1 human\n\nA determinant of size $n$ is a sum of $n!$ products, and that number grows very fast.\nHadamard's inequality shows that the determinant itself grows much more slowly, like $n^{n/2}$.\nThis is why the Fredholm determinant, a power series built from such determinants divided by $n!$, converges for every value of the parameter.\n\n### L2 human\n\nFredholm needed exactly this estimate to show that his determinant and minors are entire functions of $\\lambda$.\nWithout it, his formulas would only work near $\\lambda = 0$, where the Neumann series already does the job.\n",
  "commit": "ff8a40749dc16545e3d2fc637f6b0196e81d0bfa",
  "front": {
    "about": [
      "PPX0AK81QV"
    ],
    "kind": "statement",
    "msc": [
      "45B05",
      "15A45"
    ],
    "name": "Hadamard's inequality in Fredholm theory",
    "role": "lemma",
    "sources": [
      {
        "url": "https://en.wikipedia.org/wiki/Hadamard%27s_inequality"
      }
    ],
    "tag": "9NZ9CXC8FY",
    "uses": [
      "CGDN1X32NQ",
      "ZM5G144PZG",
      "PPX0AK81QV",
      "HP5VJ9EBST",
      "0G4F0XB9X2"
    ]
  },
  "kind": "statement",
  "link": "https://tamnd.github.io/math2/node/9NZ9CXC8FY/",
  "links": [
    {
      "field": "about",
      "target": "PPX0AK81QV"
    },
    {
      "field": "uses",
      "target": "CGDN1X32NQ"
    },
    {
      "field": "uses",
      "target": "ZM5G144PZG"
    },
    {
      "field": "uses",
      "target": "PPX0AK81QV"
    },
    {
      "field": "uses",
      "target": "HP5VJ9EBST"
    },
    {
      "field": "uses",
      "target": "0G4F0XB9X2"
    }
  ],
  "markdown": "https://tamnd.github.io/math2/node/9NZ9CXC8FY.md",
  "msc": [
    "45B05",
    "15A45"
  ],
  "name": "Hadamard's inequality in Fredholm theory",
  "prov": null,
  "role": "lemma",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/9NZ/9NZ9CXC8FY.md",
  "status": null,
  "tag": "9NZ9CXC8FY",
  "uri": "m2://9NZ9CXC8FY",
  "url": "https://tamnd.github.io/math2/wiki/hadamards-inequality-in-fredholm-theory/",
  "writing": {
    "tag": "9NZ9CXC8FY",
    "kind": "statement",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      },
      {
        "what": "a source with a locator",
        "have": 0,
        "want": 1
      }
    ]
  }
}